Question: Graph this system of equations and solve. $-x+y = 1$ $12x-2y = 8$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Explanation: Convert the first equation, $-x+y = 1$ , to slope-intercept form. $y = x + 1$ The y-intercept for the first equation is $1$ , so the first line must pass through the point $(0, 1)$ The slope for the first equation is $1$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $1$ position to the right. $1$ position to the right. $1$ position up from $(0, 1)$ is $(1, 2)$ Graph the blue line so it passes through $(0, 1)$ and $(1, 2)$ Convert the second equation, $12x-2y = 8$ , to slope-intercept form. $y = 6 x - 4$ The y-intercept for the second equation is $-4$ , so the second line must pass through the point $(0, -4)$ The slope for the second equation is $6$ . Remember that the slope tells you rise over run. So in this case for every $6$ positions you move up $1$ position to the right. $6$ positions up from $(0, -4)$ is $(1, 2)$ Graph the green line so it passes through $(0, -4)$ and $(1, 2)$ The solution is the point where the two lines intersect. The lines intersect at $(1, 2)$.